0.3 grams of platinum chloride of organic diacidic base left 0.09 grams of platinum on ignition,
the molecular weight of the organic base is

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120

240

180

answer is B.

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Detailed Solution

$2 B + H_{4} p t C l_{6} \to B_{2} H_{2} p t C l_{6} \overset{Δ}{\to} p t$

Equivalent weight of $B_{2} H_{2} p t C l_{6}$

= 2B + 410

$\frac{0.3}{2 B + 410} = \frac{0.09}{195}$

$\Rightarrow$ 3.33 × 195 = 2B + 410

239.93 = 2B

B = 120

Molecular weight of base = 2 × E. Wt

= 2 × 120

= 240

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